JEE Mathematics 2024 Question with Solution

Given:

• $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$
• $\vec{b} = 3(\hat{i} - \hat{j} + \hat{k})$
• $\vec{a} \times \vec{c} = \vec{b}$
• $\vec{a} \cdot \vec{c} = 3$

Find: $\vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{b} \cdot \vec{c}$

Use the scalar triple product identity:

$$\vec{a} \cdot (\vec{c} \times \vec{b}) = (\vec{a} \times \vec{c}) \cdot \vec{b}$$

Since $\vec{a} \times \vec{c} = \vec{b}$,

$$\vec{a} \cdot (\vec{c} \times \vec{b}) = \vec{b} \cdot \vec{b} = |\vec{b}|^2$$

Now,

$$\vec{b} = 3(\hat{i} - \hat{j} + \hat{k})$$

So,

$$|\vec{b}|^2 = 3^2\left(1^2 + (-1)^2 + 1^2\right) = 9(3) = 27$$

Next, from the given condition,

$$\vec{a} \cdot \vec{c} = 3$$

Also,

$$\vec{b} \cdot \vec{c} = (\vec{a} \times \vec{c}) \cdot \vec{c} = 0$$

because a cross product is perpendicular to $\vec{c}$.

Therefore,

$$\vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{b} \cdot \vec{c} = 27 - 0 = 27$$

Thus, the working shown in the first solution block is inconsistent with the stated answer, and the second solution block corresponds to a different expression. The source solution concludes option B, but the displayed algebra for the asked expression gives $27$, which does not match any option.

Given: $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$

Find: $\vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{b} \cdot \vec{c}$

Directly use

$$\vec{a} \cdot (\vec{c} \times \vec{b}) = (\vec{a} \times \vec{c}) \cdot \vec{b} = \vec{b} \cdot \vec{b} = |\vec{b}|^2$$

and

$$\vec{b} \cdot \vec{c} = (\vec{a} \times \vec{c}) \cdot \vec{c} = 0$$

Now,

$$|\vec{b}|^2 = |3(\hat{i} - \hat{j} + \hat{k})|^2 = 9(1+1+1) = 27$$

Hence,

$$\vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{b} \cdot \vec{c} = 27$$

the solution's marks option B as correct, but that does not agree with the displayed question expression.